7.6 problem Problem 4(b)

Internal problem ID [12381]

Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A. Dobrushkin. CRC Press 2015
Section: Chapter 8.3 Systems of Linear Differential Equations (Variation of Parameters). Problems page 514
Problem number: Problem 4(b).
ODE order: 1.
ODE degree: 1.

Solve \begin {align*} x^{\prime }\left (t \right )&=2 x \left (t \right )+6 y-2 z \left (t \right )+50 \,{\mathrm e}^{t}\\ y^{\prime }&=6 x \left (t \right )+2 y-2 z \left (t \right )+21 \,{\mathrm e}^{-t}\\ z^{\prime }\left (t \right )&=-x \left (t \right )+6 y+z \left (t \right )+9 \end {align*}

Solution by Maple

Time used: 0.079 (sec). Leaf size: 102

dsolve([diff(x(t),t)=2*x(t)+6*y(t)-2*z(t)+50*exp(t),diff(y(t),t)=6*x(t)+2*y(t)-2*z(t)+21*exp(-t),diff(z(t),t)=-x(t)+6*y(t)+z(t)+9],singsol=all)
 

\begin{align*} x \left (t \right ) &= -6 \,{\mathrm e}^{-t}+c_{3} {\mathrm e}^{6 t}+{\mathrm e}^{-4 t} c_{1} +\frac {2 c_{2} {\mathrm e}^{3 t}}{5}+12 \,{\mathrm e}^{t}-1 \\ y \left (t \right ) &= {\mathrm e}^{-t}+c_{3} {\mathrm e}^{6 t}-\frac {2 \,{\mathrm e}^{-4 t} c_{1}}{3}+\frac {2 c_{2} {\mathrm e}^{3 t}}{5}+2 \,{\mathrm e}^{t}-1 \\ z \left (t \right ) &= -6 \,{\mathrm e}^{-t}+c_{3} {\mathrm e}^{6 t}+{\mathrm e}^{-4 t} c_{1} +c_{2} {\mathrm e}^{3 t}+37 \,{\mathrm e}^{t}-4 \\ \end{align*}

Solution by Mathematica

Time used: 0.2 (sec). Leaf size: 213

DSolve[{x'[t]==2*x[t]+6*y[t]-2*z[t]+50*Exp[t],y'[t]==6*x[t]+2*y[t]-2*z[t]+21*Exp[-t],z'[t]==-x[t]+6*y[t]+z[t]+9},{x[t],y[t],z[t]},t,IncludeSingularSolutions -> True]
                                                                                    
                                                                                    
 

\begin{align*} x(t)\to -6 e^{-t}+12 e^t+\frac {3}{5} (c_1-c_2) e^{-4 t}+\frac {1}{15} (16 c_1+9 c_2-10 c_3) e^{6 t}-\frac {2}{3} (c_1-c_3) e^{3 t}-1 \\ y(t)\to e^{-t}+2 e^t-\frac {2}{5} (c_1-c_2) e^{-4 t}+\frac {1}{15} (16 c_1+9 c_2-10 c_3) e^{6 t}-\frac {2}{3} (c_1-c_3) e^{3 t}-1 \\ z(t)\to -6 e^{-t}+37 e^t+\frac {3}{5} (c_1-c_2) e^{-4 t}+\frac {1}{15} (16 c_1+9 c_2-10 c_3) e^{6 t}-\frac {5}{3} (c_1-c_3) e^{3 t}-4 \\ \end{align*}